Torque measures a force's tendency to rotate an object about an origin. A force through the pivot has no lever arm and therefore no torque about that pivot.
Change torque and moment of inertia. The same torque produces less angular acceleration for a harder-to-rotate body.
Torque 6 N·m; moment of inertia 2 kg·m²; angular acceleration 3.00 rad/s²; rotational energy at 3 rad/s 9.00 J.
The lever arm is the vector from the selected pivot to the force application point. In three dimensions:
\[\boldsymbol\tau=\mathbf r\times\mathbf F, \qquad |\boldsymbol\tau|=rF\sin\phi.\]The right-hand rule gives direction. A change of pivot changes (\mathbf r), so it can change torque even though the physical force is unchanged. In a fixed-axis two-dimensional model, project the vector torque onto the chosen axis and retain its sign; do not replace a vector cross product with a distance-only calculation unless the perpendicular geometry was already established.
Write a pure function taking pivot, application point, and force vector. Return torque components with units of newton metres. Keep force and position in one declared coordinate frame. A diagram may draw an arrow in the site blue, but it must be downstream of the same vector values and have a text result for nonvisual inspection.
Use geometric fixtures: a force through the pivot gives zero; a force parallel to the lever arm gives zero; a perpendicular force has magnitude (rF); reversing force reverses torque; and translating pivot and application point together preserves torque. Then test the rotational update separately: at fixed positive inertia, doubling signed torque doubles angular acceleration.
If a force is distributed over a body, sum or integrate its local torques about the same origin. In three dimensions, inertia is generally a tensor, so angular acceleration need not be parallel to torque. A scalar (I) lesson is only a constrained fixed-axis model; state the axis, supports, and excluded reactions before using it.
Prediction: Doubling torque doubles angular acceleration at fixed inertia.
Change torque from 3 to 6 N·m and compare the status. Then state the cross-product condition that makes torque zero.Choose pivot and coordinates first, calculate torque with one vector cross product, and test geometry separately from the fixed-axis dynamics update before connecting it to angular acceleration.
Treat the lesson as a small function before treating it as a fact to memorize. Give every value a unit, keep only the state needed for the next step, and make the output easy to inspect.
\[\text{observable output} = f(\text{inputs},\,\text{state})\]For Torque as Cross Product, write down the quantities you can control, the values your program must retain, and the result a reader could inspect. In Rotational Motion, the useful program is not the drawing: it is the smallest explicit model that makes a prediction you can test.
Prediction: changing one declared input while holding the others fixed should change only the outputs that the model connects to that input. Choose one input, predict the direction of change, then check a limiting case such as zero, a symmetric arrangement, or a familiar low-speed or small-change approximation.
This lesson is a teaching model, not a complete simulator. Before using it outside the stated question, check which interactions, scales, uncertainties, boundary conditions, and measurement limits it leaves out.
Treat Torque as Cross Product as a contract: named inputs and units enter a rule, the rule produces an observable result, and a known limit or invariant checks whether the implementation deserves trust.
Connect lever arm and force direction to torque, then use the rotational update rule to predict angular acceleration.